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Aleonysh [2.5K]
2 years ago
8

A consumer group wants to know if an automobile insurance company with thousands of customers has an average insurance payout fo

r all their customers that is greater than $500 per insurance claim. They know that most customers have zero payouts and a few have substantial payouts. The consumer group collects a random sample of 18 customers and computes a mean payout per claim of $579.80 with a standard deviation of $751.30.
Is it appropriate for the consumer group to perform a hypothesis test for the mean payout of all customers?
A. Yes, it is appropriate because the population standard deviation is unknown.
B. Yes, it is appropriate because the sample size is large enough, so the condition that the sampling distribution of the sample mean be approximately normal is satisfied.
C. No, it is not appropriate because the sample is more than 10 percent of the population, so a condition for independence is not satisfied.
D. No, it is not appropriate because the standard deviation is greater than the mean payout, so the condition that the sampling distribution of the sample mean be approximately normal is not satisfied.
E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.
Mathematics
1 answer:
zhenek [66]2 years ago
7 0

Answer:

E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

Standard deviation larger than the sample mean means that the distribution is skewed.

By the Central Limit Theorem, when the distribution is skewed, normality is assumed for samples sizes of 30 or higher. In this question, the sample is of 18, which is less than 30, so the hypothesis test is not appropriated, and the correct answer is given by option E.

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The the length of a rectangle is 9 centimeters less than four timesfour times its width. its area is 28 square centimeters. find
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Write the eguation of the line through (-8,-6) with slope = -5
Marta_Voda [28]

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Y=-5x-46.

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Read 2 more answers
Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce
34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

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Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

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b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

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3 years ago
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